Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{2n - 2}{n + 10} \times \dfrac{n^2 + 14n + 40}{n^2 + 4n} $
Explanation: First factor the quadratic. $r = \dfrac{2n - 2}{n + 10} \times \dfrac{(n + 10)(n + 4)}{n^2 + 4n} $ Then factor out any other terms. $r = \dfrac{2(n - 1)}{n + 10} \times \dfrac{(n + 10)(n + 4)}{n(n + 4)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 2(n - 1) \times (n + 10)(n + 4) } { (n + 10) \times n(n + 4) } $ $r = \dfrac{ 2(n - 1)(n + 10)(n + 4)}{ n(n + 10)(n + 4)} $ Notice that $(n + 4)$ and $(n + 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 2(n - 1)\cancel{(n + 10)}(n + 4)}{ n\cancel{(n + 10)}(n + 4)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $r = \dfrac{ 2(n - 1)\cancel{(n + 10)}\cancel{(n + 4)}}{ n\cancel{(n + 10)}\cancel{(n + 4)}} $ We are dividing by $n + 4$ , so $n + 4 \neq 0$ Therefore, $n \neq -4$ $r = \dfrac{2(n - 1)}{n} ; \space n \neq -10 ; \space n \neq -4 $